# How much would one pay for the max of two stocks?

I'm trying to figure out if stochastic calculus is the right approach for this problem... but I only vaguely understand it and I am trying to gauge if I need to spend the time learning measure theory etc etc for something that could be much more trivial.

I am going to make some simplifications here (e.g. normal instead of lognormal), so please bear with me. I'm just trying to get my bearings.

Let's say I have two stocks, X and Y.

Stock X is worth 100 and the price is normally distributed with $\sigma_X = 1$ per period.

Stock Y is worth 99 and the price is normally distributed with $\sigma_Y = 3.3333$ per period.

The two stocks are 90% correlated and the beta is $0.9 \cdot 3.333/1 = 3$

So what if I said to you, in one period, I'll give you stock X or stock Y, whichever has a greater value.

So the question is... what's the expected value of the max of the two stocks?

I can, of course, simulate this scenario, but is there a way to get the answer closed form?

Using Cholesky decomposition, I simulate two correlated random normals and give them the appropriate mean and standard deviation.

I approximate the value of $\max(X,Y) \approx 100.6$

• Welcome to Quantitative Finance! Would you please add your research on the matter and what you think the answer might be? (Perhaps a tree with values of starting and ending states). – rajah9 Dec 23 '15 at 20:42
• Is there a place I can upload my excel file? – TH4454 Dec 23 '15 at 20:45

Consider two jointly normal random variables $X_1 \sim N(u_1, \sigma_1^2)$ and $X_2 \sim N(u_2, \sigma_2^2)$. Note that, \begin{align*} \max(X_1, X_2) = X_2 + \max(X_1-X_2, \ 0). \end{align*} Moreover, $X_1-X_2$ is a normal random variable with mean $\mu=\mu_1-\mu_2$ and variance \begin{align*} \sigma^2 = \sigma_1^2+\sigma_2^2 - 2 \rho\sigma_1\sigma_2, \end{align*} where $\rho$ is the correlation. That is, \begin{align*} X_1-X_2 = \mu + \sigma \xi, \end{align*} where $\xi$ is a standard normal random variable. Therefore, \begin{align*} E\left(\max((X_1, X_2) \right) &= E(X_2) + E\left(\max(\mu + \sigma \xi, \ 0)\right)\\ &=\mu_2 + \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\max(\mu + \sigma x, \ 0)\,e^{-\frac{x^2}{2}}\,dx\\ &=\mu_2 + \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}(\mu + \sigma x) \,\mathbb{I}_{\mu + \sigma x\geq 0}\,e^{-\frac{x^2}{2}}\, dx\\ &=\mu_2 + \frac{1}{\sqrt{2\pi}}\int_{-\frac{\mu}{\sigma}}^{\infty}(\mu + \sigma x)\,e^{-\frac{x^2}{2}} dx\\ &=\mu_2 + \mu \,N\Big(\frac{\mu}{\sigma}\Big) +\frac{\sigma}{\sqrt{2\pi}}\,e^{-\frac{\mu^2}{2\sigma^2}}, \end{align*} where $N$ is the cumulative distribution function of a standard normal random variable.

• The end result gives me the right answer, but how did you jump from E(max(mu + sigma*epsilon,0) to the integral? – TH4454 Dec 28 '15 at 16:48
• I added a few more steps. – Gordon Dec 28 '15 at 17:35

This took a 5 second Google search: https://stats.stackexchange.com/questions/139072/distribution-of-the-maximum-of-two-correlated-normal-variables

All you're really looking is simply to compute $E[\max\{X, Y\}]$, where $X, Y$ are two correlated normal random variables. The link there tells you the pdf of the random variable $W = \max\{X,Y\}$. From there, you can integrate and compute $E[W]$ --- although most likely you can only numerically integrate in this case.

Remark: Furthermore, if I may add, there's nothing "stochastic calculus" in your problem (at least the way that you phrase it). And also, measure theory is also unnecessary for this problem.

• That gives the PDF, but not the expected value. Please try not to be rude, I've given considerable thought to this question. – TH4454 Dec 23 '15 at 22:11
• Actually, I stand corrected. Looking at the reference gwern.net/docs/conscientiousness/2008-nadarajah.pdf right in the link above, there is a closed form expectation --- see equation (9) – user32416 Dec 23 '15 at 22:26
• The paper has equations (9) and (10) for the first two moments. I can just use those for expected value and variance of the max(X,Y), right? – TH4454 Dec 23 '15 at 22:27
• Yes. And actually with the mgf, you can get any moments. Please accept my answer if you think this solves your problem. – user32416 Dec 23 '15 at 22:28

I'd recommend following Margrabe's formula, which is a closed form Black-Scholes type formula for valuing european call option whose payoff at time T = max[0, S2(T) - S1(T)]. If you create a portfolio of S1 and such a call option, your payoff will be higher of the two stock prices - which is what you want. Plus this has a more general assumption of prices being log-normally distributed.

Margrabe's call option formula: C = S2.N(d1) - S1.k.N(d2)

For d1, d2 and other details please refer to the paper: http://www.stat.nus.edu.sg/~stalimtw/MFE5010/PDF/margrabe1978.pdf